We have,

$\frac{3+2 i \sin \theta}{1-2 i \sin \theta} =\frac{(3+2 i \sin \theta)(1+2 i \sin \theta)}{(1-2 i \sin \theta)(1+2 i \sin \theta)} $

$=\frac{3+6 i \sin \theta+2 i \sin \theta-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}=\frac{3-4 \sin ^{2} \theta}{1+4 \sin ^{2} \theta}+\frac{8 i \sin \theta}{1+4 \sin ^{2} \theta}$

We are given the complex number to be real. Therefore

$\frac{8 \sin \theta}{1+4 \sin ^{2} \theta}=0, \text { i.e., } \sin \theta=0$

Thus $\quad \theta=n \pi, n \in Z$